If the straight line ax+by=2 ;a ,b 0 touches the circle x^2+y^2-2x=3 and is normal to the circle x^2+y^2-4y=6then the value of b is

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Correct answer: 1. 1JEE Main 2020 Given, x^2+y^2-2x=3 ∴ Centre is (1 ,0 ) and radius is 2 and x^2+y^2-4y=6 ∴ Centre is (0 ,2 ) and radius is 10. Since line ax+ by=2 touches the first circle ∴ a (1 )+b (0 )-2 a^2+b^2=2 or (a-2)= (2 a^2+b^2 ) … (i ) Also, the given line is normal to the second circle. Hence it will pass through the centre of thesecond circle. ∴a (0 )+b (2 )=2 or 2b=2 or b=1 Putting this value in equation (i ) we get a-2=2 a^2+1^2 or (a-2 )^2=4 (a^2+1 ) or a^2+4-4a=4a^2+4or 3a^2+4a=0 or a (3a+4 )=0 or a=0 ,- 43 ∴ Values of a and b are (- 43 ,1) respectively according to the given choices

$If the straight line a x + b y = 2; a, b \neq = 0 touches the circle x^{2} + y^{2} - 2 x = 3 and is normal to the circle x^{2} + y^{2} - 4 y = 6 then the value of b is$

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